Prime Ring Problem

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

 

Input

n (0 < n < 20).

 

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

 

Sample Input

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;static int j = 1;int n, a[20];bool prime[40], vis[20];bool is_prime(int m) {    if(m == 1)        return false;    for(int i = 2; i*i <= m; i++)        if(m % i == 0) return false;    return true;}void dfs(int cur) {    if(cur == n && prime[a[0]+a[n-1]]) {        for(int i = 0; i < n; i++) {            cout << a[i];            if(i < n-1)                cout << ' ';            else                cout << endl;        }    }    else for(int i = 2; i <= n; i++)        if(!vis[i] && prime[i+a[cur-1]]) {            a[cur] = i;            vis[i] = 1;            dfs(cur+1);            vis[i] = 0;        }}int main(){    memset(a, 0, sizeof(a));    memset(prime, false, sizeof(prime));    memset(vis, false, sizeof(vis));    for(int i = 1; i < 40; i++)        prime[i] = is_prime(i);    while(scanf("%d", &n) != EOF) {        a[0] = 1;        printf("Case %d:\n", j++);        dfs(1);        cout << endl;    }    return 0;}